The intensity of central fringe in the interference pattern produced by two indetical slits is I. When one of the slits is closed then the intensity at the same points is `I_(0)`. The relation between I and `I_(0)` is

Options

A. `I = 4I_(0)`
B. `I = 2I_(0)`
C. `I = I_(0)`
D. `I = (I_(0))/(2)`

Answer

To solve the problem, we need to establish the relationship between the intensity of the central fringe (I) in the interference pattern produced by two identical slits and the intensity (I₀) when one of the slits is closed.

### Step-by-Step Solution:

1. **Understanding the Intensity in Interference:**
- When both slits are open, the intensity at the central fringe (I) is given by the formula:
\[ I = (\sqrt{I_1} + \sqrt{I_2})^2 \]
- Here, I₁ and I₂ are the intensities from each slit. Since the slits are identical, we have:
\[ I_1 = I_2 = I_0 \]

2. **Calculating the Intensity with Both Slits Open:**
- Substituting I₁ and I₂ into the intensity formula:
\[ I = (\sqrt{I_0} + \sqrt{I_0})^2 \]
- This simplifies to:
\[ I = (2\sqrt{I_0})^2 = 4I_0 \]

3. **Understanding the Intensity with One Slit Closed:**
- When one of the slits is closed, the intensity at the same point becomes:
\[ I_0 = I_1 = I_0 \]
- Thus, the intensity when one slit is closed is simply I₀.

4. **Establishing the Relationship:**
- From the calculations, we have established that:
\[ I = 4I_0 \]
- Therefore, the relationship between I and I₀ is:
\[ I = 4I_0 \]

### Final Answer:
The relationship between I and I₀ is:
\[ I = 4I_0 \]